How does motor current increase with constant torque but increasing HP

[mull982]

I am having trouble understanding how current in a motor increases when the toruqe demand on the motor stays constant, however the HP requirement increases.

For something like a constant torque load if the load is kept the same but the load increases in speed somehow (vfd, etc..) then I understand that the torque stays the same but yet the hp increases and therefore the current increases.

I was under the assumption that the current drawn by a motor was directly propotional to the slip on the motor which in turn dictated the toruqe on the motor. By increaseing the slip at a given speed the torque was increased, and therefore the current was increases. But if the torque requirement is staying the same then in theory how is the current increasing in the motor.

The motor has to increase in current due to the fact that the HP increases and HP = V * I * 1.73 *pf. So what happens inside the motor when the torque stays the same but the hp changes as a result of increased speed.

 

[One-eyed Jack]

As the speed of any given motor goes up the available torque go down. The hp remains constant. 1hp = 3ft lb of torque at 1800 rpm 1hp=1.5 ft lb @ 3600 rpm

 

[winnie]

As a _rough_ approximation, for any given motor, current is proportional to torque and _voltage_ is proportional to speed.

A somewhat better approximation is that for any given motor, magnetic flux times current is proportional to torque. If you keep magnetic flux _constant_ then current is proportional to torque. To maintain a fixed level of magnetic flux, voltage is proportional to rotating field speed (drive frequency).

So if you are running a motor with a VFD, and supplying a constant torque load, and if you change drive frequency and voltage to maintain a constant V/f, then I would expect the current to the motor to remain constant, but the voltage to change with speed. Voltage times current remains proportional to torque times speed.

If on the other hand you increase frequency _without_ increasing voltage, then the magnetic flux will drop, and you will need more current to supply the same torque. Voltage times current remains proportional to torque times speed, but in this case you are keeping voltage constant so the current needs to change.

-Jon

 

[Besoeker]

I was under the assumption that the current drawn by a motor was directly propotional to the slip on the motor which in turn dictated the toruqe on the motor.

It's the rotor current that is proportional to torque. Stator current might be around 30% of FLC even at no load, so it is not directly proportional to torque.

1. Torque is the product of stator flux and rotor current.
2. Mechanical power (HP) is the product of torque and rotational speed.
3. From #2, a 10% increase in speed for the same torque results in a 10% increase in mechanical power.
4. To get the 10% speed increase with a VFD, you would usually increase both output voltage and frequency by 10% thus keeping the same V/f ratio.
5. If the V/f ratio stays the same, the flux stays the same.
6. Back to point #1. Same flux, same torque, so the same current. The output power has gone up, the current has not.

 

[mull982]

It's the rotor current that is proportional to torque. Stator current might be around 30% of FLC even at no load, so it is not directly proportional to torque.

1. Torque is the product of stator flux and rotor current.
2. Mechanical power (HP) is the product of torque and rotational speed.
3. From #2, a 10% increase in speed for the same torque results in a 10% increase in mechanical power.
4. To get the 10% speed increase with a VFD, you would usually increase both output voltage and frequency by 10% thus keeping the same V/f ratio.
5. If the V/f ratio stays the same, the flux stays the same.
6. Back to point #1. Same flux, same torque, so the same current. The output power has gone up, the current has not.

Thanks! Very clear example! I didn't think of the whole V/Hz ratio changing before thus decreasing voltage for increased speeds.

That was a great example for a linear HP increase. But what about for an application such as a fan whose HP changes with the cube of the speed? I dont think a simple linear V/Hz change will maintain the same current will it? I guess this is sort of a tough one since the tourque would also be changing as well.

 

[Besoeker]

Thanks! Very clear example!

Thank you for your kind words.

But what about for an application such as a fan whose HP changes with the cube of the speed? I dont think a simple linear V/Hz change will maintain the same current will it? I guess this is sort of a tough one since the tourque would also be changing as well.[/QUOTE]
Power, as you say, changes with the cube of the speed for a centrifugal load. Torque changes as the square of the speed. For fixed V/f ratio, the rotor current will also go up with the square of the speed.

The stator current will also go up but the relationship with torque is a little more complex and depends on the characteristics of the particular motor.

This for an installation we did a while back. There were three 625 kW 660V motors driving centrifugal pumps. Speed range is 70% to 100% hence the bottom end torque is 0.49 pu or 49%.
A couple of points to note. Maximum duty is at 50Hz and I have scaled the axes to show operating range only.

 

[philly]

Thank you for your kind words.

But what about for an application such as a fan whose HP changes with the cube of the speed? I dont think a simple linear V/Hz change will maintain the same current will it? I guess this is sort of a tough one since the tourque would also be changing as well.

Power, as you say, changes with the cube of the speed for a centrifugal load. Torque changes as the square of the speed. For fixed V/f ratio, the rotor current will also go up with the square of the speed.

The stator current will also go up but the relationship with torque is a little more complex and depends on the characteristics of the particular motor.

This for an installation we did a while back. There were three 625 kW 660V motors driving centrifugal pumps. Speed range is 70% to 100% hence the bottom end torque is 0.49 pu or 49%.
A couple of points to note. Maximum duty is at 50Hz and I have scaled the axes to show operating range only.

[/QUOTE]

Great examples using VFD's. Many applications on our plant are directly driven or use a gearbox. The question comes up time and again as to what happens when you replace the same HP motor with different speeds in these applications.

For instance if I have a 100hp 1800rpm motor directly driving a conveyor what happens when I put a 100hp 1200rpm motor in its place? Since the speed goes down, then the rated output torque of the motor must go up. But how is this, since torque is a function of current, and the current will be roughly the same for any given 100hp motor. In other words, how is torque changing in this instance but current staying the same.

So for this example if we had a constant torque load such as a conveyor and we went from this 1800rpm motor to a 1200rpm motor then we would be reducing the hp requirement of the motor because of the decrease in speed. However since the torque stays constant, we would expect the current to stay constant. How in this case then are we reducing required hp from load but reducing current if torque is staying the same?

 

[Besoeker]

Great examples using VFD's. Many applications on our plant are directly driven or use a gearbox. The question comes up time and again as to what happens when you replace the same HP motor with different speeds in these applications.

For instance if I have a 100hp 1800rpm motor directly driving a conveyor what happens when I put a 100hp 1200rpm motor in its place? Since the speed goes down, then the rated output torque of the motor must go up. But how is this, since torque is a function of current, and the current will be roughly the same for any given 100hp motor. In other words, how is torque changing in this instance but current staying the same.

As you rightly say, the rated output torque of the 1200rpm motor is greater than that of the 1800 rpm motor. But, for a constant torque load, the required load torque remains the same, regardless of speed.
So, even though the 1200 rpm motor can produce about 50% more torque than the 1800 rpm machine it has to produce only the same. So it will run at a lower percentage of its rated torque and thus a lower percentage of it's rated full load current.

I have gone that through a couple of times and I still don't think it reads altogether well. But maybe you can the gist of it.

 

[philly]

As you rightly say, the rated output torque of the 1200rpm motor is greater than that of the 1800 rpm motor. But, for a constant torque load, the required load torque remains the same, regardless of speed.
So, even though the 1200 rpm motor can produce about 50% more torque than the 1800 rpm machine it has to produce only the same. So it will run at a lower percentage of its rated torque and thus a lower percentage of it's rated full load current.

I have gone that through a couple of times and I still don't think it reads altogether well. But maybe you can the gist of it.

This makes sense. The only thing that is puzzling is how for the same full load current, we can have two different full load torques.

For example with the 100hp 1800rpm motor this will have a torque rating of 291.6 ft-lbs. For the same 100hp motor with a 1200rpm speed the torque rating will be 437.5 ft-lbs. However both of these motors will still have a full load current of aprox 125A. So how can these motors have the same full load current but two different full load torques. I thought with the V/Hz ratio constant then the output torque was equivelent to the current as mentioned above which would mean that for the same current we should see the same toruqe. Am I missing something?

 

[winnie]

This makes sense. The only thing that is puzzling is how for the same full load current, we can have two different full load torques.

For example with the 100hp 1800rpm motor this will have a torque rating of 291.6 ft-lbs. For the same 100hp motor with a 1200rpm speed the torque rating will be 437.5 ft-lbs. However both of these motors will still have a full load current of aprox 125A.

But these cannot be the _same_ motors. They are two different motors with two different windings, two different iron lamination designs, two different magnetic pole counts, etc.

Once you change the design of the motor, all bets are off as to the proportionality between torque and current.

-Jon

 

[Besoeker]

This makes sense.

It did??! We had a full house with a bunch of the overseas outlaws, all us talking at the time and, being the holiday season, just possibly the spirit of goodwill was upon us...:grin:
OK. A bit more seriously......

The only thing that is puzzling is how for the same full load current, we can have two different full load torques.

For example with the 100hp 1800rpm motor this will have a torque rating of 291.6 ft-lbs. For the same 100hp motor with a 1200rpm speed the torque rating will be 437.5 ft-lbs. However both of these motors will still have a full load current of aprox 125A. So how can these motors have the same full load current but two different full load torques.

I'll try.

1. Both motors are rated at 100 hp output.
2. To keep it to basics, assume that both have the same efficiency and power factor.
3. From point #2, both motors will have the same input current and power at 100 hp output.
4. Output mechanical power is torque times rpm.
5. To get the same 100 hp output, the slower machine has to produce more torque.

An analogy.
Use the 1800 rpm (4-pole) motor. Stick an 1800:1200 ratio reduction gearbox on its output shaft and you reduce the output speed, increase the output torque. But you get the same output power.
Using a 6-pole 1200 rpm motor does just that. It's like changing a gearbox ratio.

 

[Besoeker]

I mentioned gear ratios in the context of this thread.
I hope the mods and the admin will allow this slightly off-topic post to go - season of goodwill. Power, gear ratios and electrical problems are all mentioned....

I used to cycle quite a lot. I lived close enough to my office that it was a practical proposition to use my bike.
It had 18 different gear ratios to select from. This was quite useful given that I had a steep slope to descend going to my office - and had to climb on my return. Same input power, just me, but different output torques.

One Saturday evening I got a call from a customer with an electrical problem. I didn't have the drawings at home so I agreed to go to my office, get out the drawings, and talk him through the problems. Better than the 500 mile drive to his site.

The steep slope that had previously been pretty rough and a slow descent have been paved and as smooth as a pool table. Brilliant I thought. Maybe 35 mph down the slope. Until I hit the speed hump. I regained consciousness some time later to see around me a crowd, an ambulance, and some blood running down the road.

It was still in my mind that I had a customer requiring support so I dismissed the ambulance crew and went on to my office. The bike was too damaged to ride so I carried it. It was less than a couple of miles.

I got the drawings out, we resolved the problem. I apologised for the rather slow response. They had called me a couple of times on my cell phone but my hands were too damaged to get it out of my pocket.
But job done and a happy customer.
Support keeps customers.

My brain finally took some note of my own physical condition.
I was in some pain and bleeding like a stuck pig. I tried to get an ambulance to get me to the local hospital but they were all busy. So I walked there. Just another couple of miles and no bike to carry. No big deal.

They fixed me up. A few stitches for the gash in my head, a sling to support my dislocated shoulder and some pain medication. The broken collar bone was just collateral damage and expected to knit together in time.
And I was sent on my way. No transport so I walked home. Several miles.
Sometimes you just have to go that extra mile for valued customers.

 

[philly]

I mentioned gear ratios in the context of this thread.
I hope the mods and the admin will allow this slightly off-topic post to go - season of goodwill. Power, gear ratios and electrical problems are all mentioned....

I used to cycle quite a lot. I lived close enough to my office that it was a practical proposition to use my bike.
It had 18 different gear ratios to select from. This was quite useful given that I had a steep slope to descend going to my office - and had to climb on my return. Same input power, just me, but different output torques.

One Saturday evening I got a call from a customer with an electrical problem. I didn't have the drawings at home so I agreed to go to my office, get out the drawings, and talk him through the problems. Better than the 500 mile drive to his site.

The steep slope that had previously been pretty rough and a slow descent have been paved and as smooth as a pool table. Brilliant I thought. Maybe 35 mph down the slope. Until I hit the speed hump. I regained consciousness some time later to see around me a crowd, an ambulance, and some blood running down the road.

It was still in my mind that I had a customer requiring support so I dismissed the ambulance crew and went on to my office. The bike was too damaged to ride so I carried it. It was less than a couple of miles.

I got the drawings out, we resolved the problem. I apologised for the rather slow response. They had called me a couple of times on my cell phone but my hands were too damaged to get it out of my pocket.
But job done and a happy customer.
Support keeps customers.

My brain finally took some note of my own physical condition.
I was in some pain and bleeding like a stuck pig. I tried to get an ambulance to get me to the local hospital but they were all busy. So I walked there. Just another couple of miles and no bike to carry. No big deal.

They fixed me up. A few stitches for the gash in my head, a sling to support my dislocated shoulder and some pain medication. The broken collar bone was just collateral damage and expected to knit together in time.
And I was sent on my way. No transport so I walked home. Several miles.
Sometimes you just have to go that extra mile for valued customers.

Wow what a story and dedication. I appreciate you going the extra mile as well on this forumn to help answer all of the questions I have had.

It did??! We had a full house with a bunch of the overseas outlaws, all us talking at the time and, being the holiday season, just possibly the spirit of goodwill was upon us...:grin:
OK. A bit more seriously......



I'll try.

1. Both motors are rated at 100 hp output.
2. To keep it to basics, assume that both have the same efficiency and power factor.
3. From point #2, both motors will have the same input current and power at 100 hp output.
4. Output mechanical power is torque times rpm.
5. To get the same 100 hp output, the slower machine has to produce more torque.

An analogy.
Use the 1800 rpm (4-pole) motor. Stick an 1800:1200 ratio reduction gearbox on its output shaft and you reduce the output speed, increase the output torque. But you get the same output power.
Using a 6-pole 1200 rpm motor does just that. It's like changing a gearbox ratio.

I do understand the concept of HP equaling output torque times speed, it was the concept of torque values in relation to input current that I was not understanding. But it looks like as Winnie said in his response, that the torque response to input current will be dependent on the physical design of the motor including all the things he listed below.

But these cannot be the _same_ motors. They are two different motors with two different windings, two different iron lamination designs, two different magnetic pole counts, etc.

Once you change the design of the motor, all bets are off as to the proportionality between torque and current.

-Jon

So for any given motor the torque magnitude will be depended on the input current, and the physical design properties of this motor. So to change torque in a motor you must rewind to change these properties.

 

[frenchelectrican]

So for any given motor the torque magnitude will be depended on the input current, and the physical design properties of this motor. So to change torque in a motor you must rewind to change these properties.

Most case it is true expect the Sychronous motor with the Sychronous motor with mulit speed motor set up they will have switching poles so they can change the speed on the spot like example I can run in 10 pole mode then if want to slow down the speed I can go to 20 pole mode or even 16 pole mode it will change the T/min { RPM } plus toqure level.

Merci,Marc

 

[Besoeker]

So for any given motor the torque magnitude will be depended on the input current,

Well yes, but it is maybe easier to think of it the other way round.
The load torque demand determines the required motor input power.

 

[Besoeker]

For what it's worth, here is the relationship between input current and output torque for a 500 kW we provided.

 

[philly]

Well yes, but it is maybe easier to think of it the other way round.
The load torque demand determines the required motor input power.

So to help understand it with an example, lets say I have a constant torque conveyor that requires aprox 250 ft-lbs of torque and I try running this load with both the 1800rpm and 1200rpm 100hp motors.

Using the 1800rpm motor, the motor would have a full load torque of 291 ft-lbs and because the required torque load of 250 ft-lbs was close to the full load torque vaule of this motor we would expect to be drawing somewhere near full load current.

If we were to use the 1200rpm motor to drive this load then this motor would have a full load torque of 437.5 ft lbs, and becasue the required torque of 250ft-lbs was roughly a little more than half the max toruqe of this motor, we would expect to see somewhere around 50% full load current (ignoring pf and efficiency) or somewhere between 50-60A.

So ignoring the speed for a second the physical property that lets a motor have more torque for a given current input then another motor is the fact that is has a different winding configuration, winding material, rotor material etc... Is there any good source of information for understanding how these physical properties of the motor determine the torque for a given current input?

For what it's worth, here is the relationship between input current and output torque for a 500 kW we provided.

So if we were to somehow rewind this motor or find a similar hp output motor with a lower rpm then the torque curve shown would shift to the right thus allowing a greater torque output for a given input current?

 

[dkarst]

Maybe you're looking for a more in-depth technically satisfying response but here is a very simple explanation using one parameter. Let's say you have your 100HP 1800 rpm motor in parts on the table and you're trying to invent a 100HP 1200 rpm out of the parts. We first need a couple more poles to get from a 4 pole machine to a 6 pole. Assuming we can find/invent a couple more identical poles, we may need to increase the stator diameter to get them to physically "fit". In fact, we have to scale the whole machine in diameter (rotor and stator). Now one can see (assuming a whole much of things not included here) that the torque lever arm exerted on the machine's rotor is greater, since Torque = force X lever arm. Since we increased the number of poles, we may have to shorten the length of the machine to keep some of the other operating parameters the same. So now we have a 1200 rpm 6 pole machine with greater torque at the operating point than before, but the same power output. Someone will probably be along shortly with alternate explnations using number of turns in the stator etc. but I thought this made sense to get a basic understanding.

One can imagine the torque generated with a large hydrogenerator turning at ~ 100rpm and producing multi-MW which is why a huge shaft is required.

 

[Besoeker]

So to help understand it with an example, lets say I have a constant torque conveyor that requires aprox 250 ft-lbs of torque and I try running this load with both the 1800rpm and 1200rpm 100hp motors.

Using the 1800rpm motor, the motor would have a full load torque of 291 ft-lbs and because the required torque load of 250 ft-lbs was close to the full load torque vaule of this motor we would expect to be drawing somewhere near full load current.

If we were to use the 1200rpm motor to drive this load then this motor would have a full load torque of 437.5 ft lbs, and becasue the required torque of 250ft-lbs was roughly a little more than half the max toruqe of this motor, we would expect to see somewhere around 50% full load current (ignoring pf and efficiency) or somewhere between 50-60A.

That's basically correct except that you can't make a direct correlation between torque and stator current. The stator current comprises essentially two parts. First there is the magnetising current which is there and approximately constant regardless of motor loading. This might typically be around 30%-40% (see the curve I posted) of rated FLC even under no-load conditions. Then there is the reflected rotor current which is proportional to mechanical loading.

So ignoring the speed for a second the physical property that lets a motor have more torque for a given current input then another motor is the fact that is has a different winding configuration, winding material, rotor material etc...

Well, no. Specifically, you'd need a different number of pole pairs. They are always in pairs. And a lower speed machine would be physically larger.

The stator winding produces a rotating magnetic field. The rate at which it rotates depends on the number of pole pairs. On 60 Hz it will get from one pole pair to the next in 1/60 of a second. for a motor with just one pole pair, commonly called a 2-pole motor, it makes 60 rotations of the motor per second or 3600 per minute. This rotating field induces current in the rotor resulting in torque In normal operation the rotor spins at very nearly the same speed as the rotating field and the small difference, as you probably know, is called slip.
If there are more pole pairs, the time taken from each pole pair to the next is the same, but the time for a complete revolution of the motor is correspondingly longer resulting is a lower speed, higher torque motor for the same rated output.
Your 4-pole 1800 rpm and 6-pole 1200 rpm motors are examples of just that.

Is there any good source of information for understanding how these physical properties of the motor determine the torque for a given current input?

My lecturer? The guy I had for electrical machine theory was very good.
He had actually worked for a company that made motors and generators.
But that was four decades ago.

So if we were to somehow rewind this motor

Not really an option.

or find a similar hp output motor with a lower rpm then the torque curve shown would shift to the right thus allowing a greater torque output for a given input current?

The torque curve I gave is in pu (per unit) values.
It would shift. What would be different is what 1.0 pu torque was in actual terms.
The reason I used pu values was to show just the general nature of the relationship. And my actual values for torque were in kNm. I was too indolent to do the conversion........SI is good!